Take a smooth closed curve in the plane. At each self-intersection, randomly choose one of the two pieces and lift it up just out of the plane. (Perturb the curve so there are no triple intersections.) I don't really know anything about knot theory, so I don't even know if I'm asking the right questions here, but I'm wondering: What is the probability that this is the trivial knot? What can we say about how knotted this knot might be, and with what probabilities? (Measure "knottedness" in whatever way you like.) More generally, can we say anything about the probability of the various possible values in the usual invariants that people use to study knots?

I only have an idea of how to approach the first question, and even then it's only by brute force. I was just playing around with the easiest cases, and I think that with 0, 1, or 2 intersections, all knots are trivial, and with 3 intersections the knot is trivial with probability 75%.

A general analysis should presumably involve calculating the probability that we can simplify using various Reidemeister moves, but I don't know how to incorporate this. I'd imagine a computer could brute-force the first few cases pretty easily (I'm not so bold as to venture an order-of-magnitude guess on whether it's the first few hundred or the first few million)...

7 Answers
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One possible route to a model of random knots would be through the braid group. Every knot can be expressed (non-uniquely) as the closure of a braid. So, for example, you could apply the braid generators uniformly $n$ times across $k$ strands, close the braid using your favorite closure, and then ask this question sensibly. I don't think you can directly ask about the $n \to \infty$ limit for the braid group, though, because I don't think there is a notion of uniform measure for that group. Actually, perhaps I will post this as a separate question, but is the braid group amenable? I would wager that in this model, the probability of having the unknot decreases very quickly with $n$ and $k$.

To test if you have the unknot, it is conjectured that you just have to check the Jones polynomial. But even this is still hard in general, unlesseven if you happen to have a quantum computer. :)

Consider the subgroup consisting of braids where all but the last strand stand still, and the last strand winds around them. This subgroup is clearly free, being the fundamental group of an (n-1)-punctured disk, and so braid groups are not amenable.
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Tom ChurchNov 6 '09 at 23:44

As an aside, the process of "combing" a braid is given by filtering by the cosets of such subgroups. This exhibits the pure braid group as an iterated extension of free groups; this was used by Arnol'd in his beautiful computation of the cohomology ring of the pure braid group [MR242196]. (Arnold's paper is very readable, but the translation can be a bit hard to track down; for anyone who is interested, I have some handwritten notes on the cohomology of braid groups, including Arnold's proof, on my website.)
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Tom ChurchNov 6 '09 at 23:45

I suspect that answering this question would be very difficult. A more reasonable question would be to try to understand the distribution of the various numerical knot invariants. I don't know any references off hand, but I know I've heard talks on the subject.

If you want to try to make conjectures about this kind of thing, then I highly recommend Livingston's table of knot invariants, which contains an amazing amount of data.

The model you propose for random knots obviously depends on the curve you draw initially, so I'm not sure this is the most natural model to consider. People have certainly looked at various probability distributions of (various classes of) knots (or knot projections). One of the immediate problems is that even just doing computer simulations is hard since determining the knot type - or just unkottedness - of a given knot diagram is highly non-trivial.

A paper which does this with Vassiliev Invariants (a certain important class of polynomial-like invariants of knots) appears in the volume "Random Knotting and Linking", edited by Millett and Summers (look at the paper by Deguchi and Tsurusaki). Other papers in this volume may interest you, too.

To the best of my knowledge, there's no really model of random knots for which the question "what is the probability that the knot is trivial" has a known answer, except that as the number of crossing tends to infinity this probability likely approaches 0 (as anyone who left a set of mobile headphones in his pocket for more than five minutes knows).

Is it not clear which Reidemeister moves one should apply to simplify a knot? I guess I'd imagine it could be that it needs to get more complicated before it can get any simpler.
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Aaron Mazel-GeeOct 30 '09 at 5:59

Yes, that is correct: it is not clear which move to apply to a given not diagram in order to simplify it (whatever "simplify" means), and some trivial knot diagrams have the property that they cannot be reduced to the unknot without introducing some additional intersections first. If this hadn't been the case, this whole beautiful theory would have been reduced to a simple algorithm...
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Alon AmitOct 30 '09 at 6:09

In general, deciding which Reidemeister moves to do is a very difficult problem. While algorithms have been known for a long time (at least since the work of Haken in the '60's), they are very complicated and not at all practical. In particular, there are examples where you have to introduce a huge number of new crossings before your knot can start to be simplified. An accessible source of examples is Kauffman-Lambropoulou's paper "Hard Unknots and Collapsing Tangles", available on the arXiv here : arxiv.org/abs/math/0601525
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Andy PutmanOct 30 '09 at 6:09

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One approach that might be interesting and which avoids hard unknots is to represent a knot as a grid diagram and see whether it admits any series of commutation moves followed by a destabilization (these are some of the grid diagram analogues of Reidemeister moves). It's not clear what such a series of moves would be or how to figure this out efficiently, but a knot is the unknot iff you can repeat this until you get the trivial 2x2 diagram. See Dynnikov's paper "Arc-presentations of links. Monotonic simplification", arXiv:0208153.
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Steven SivekOct 30 '09 at 13:17

You should look at the Knot Atlas, which contains lots of tabulated knot invariants, although often not in as convenient form as Livingston's site.

Really, though, you want to download the KnotTheory` package (presupposing you have access to Mathematica), available at the Knot Atlas. With a bit of fiddling, you can easily run experiments of the type you describe. It can calculate many invariants from the presentation of a knot.

Best of all, you should go and think about "physically realistic" models of random knots, and then try to implement such a model using one of the many knot notations the KnotTheory` package understands. There are some good papers written about this subject, and even some real life experiments with strings in boxes being shaken up and down! :-)

I believe there are a few known "random knotting" type results out there. Not the kind of results the original poster requested, but related. Take n points in R^3 generated by a random walk, join them up (cyclicly) by straight lines. That's generically a knot. And with probability 1 (as n gets large) it's non-trivial and has a trefoil knot summand. The paper by Deguchi and Tsurusaki in "Lectures at Knots '96" provides references for these results although I've never read them in detail.

SnapPea won't recognise torus knots, nor non-prime knots, or knots whose complements have incompressible tori. So if your random knots have a lot of prime summands (which is common to a lot of random knot generators), SnapPea will choke most of the time.
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Ryan BudneyNov 10 '09 at 3:16

Burton, Rubinstein, Jaco and Tillmann are getting pretty close to having efficient algorithms for recognising such knots.
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Ryan BudneyNov 10 '09 at 3:19

Sorry to disagree, but if the JSJ decomposition has a hyperbolic piece then SnapPea will present the splitting torus in the "splitting window". It does this by tracking the degeneration of the tetrahedra, and so finds the quad type (the speed of degeneration tells you the number of quads!) SnapPea can also sometimes guess at SL(2,R) representations (ie detect Seifert fibred spaces).
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Sam NeadNov 10 '09 at 3:24

Re: Burton, Rubinstein, Jaco, Tillmann. I assume that their techniques will still be at least exponential time. SnapPea is not an algorithm, but it has the virtue of being fast! There are ways to kill SnapPea (eg feed it surface bundles where the monodromy is a high power and then ask it for a Dirichlet domain), but it is pretty hard to kill SnapPea with a hand-drawn knot...
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Sam NeadNov 10 '09 at 3:30